Whether the base field is algebraically closed doesn't matter here, since two matrices are similar over a small field if and only if they are similar over a bigger field. So the question can always be reduced to an algebraically closed field.
That said, it can be shown that two matrices are similar if and only if the Jordan normal form has the same "invariant factors" and it can be shown that a matrix and it's transpose have the same invariant factors. So a matrix is similar to it's transpose.
Check out the book "matrix analysis" by Horn and Johnson.
#5
arthurhenry
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I thank you Micromass, that source was very helpful.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?